The distance between ${\mathrm{Na}}^{+}$ and ${\mathrm{Cl}}^{-}$ ions in solid NaCl of density 43.1 g ${\mathrm{cm}}^{-3}$ is _____ $\mathrm{x}{10}^{-10}$  m. (Nearest Integer) (Given: ${\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23}{\mathrm{mol}}^{-1}$)

For $$\begin{gathered} \mathrm{NaCl} \\ Z=4\text{ and }\mathrm{M}=58.5\text{ gram } \end{gathered}$$

$\begin{array}{l}\mathrm{\rho }=\frac{\mathrm{ZM}}{{\mathrm{N}}_{\mathrm{A}}\text{ Volume }}\\ 43.1=\frac{4\times 58.5}{6\times {10}^{23}\times [\mathrm{a}{]}^3}\\ {\mathrm{a}}^3=\frac{4\times 58.5}{6\times 43.1}\times {10}^{-23}=0.9\times {10}^{-23}=9\times {10}^{-24}\\ \mathrm{a}=2.08\times {10}^{-8}\mathrm{cm}\end{array}$

$\begin{array}{l}{\mathrm{d}}_{{\mathrm{Na}}^{+}+{\mathrm{cl}}^{-}}=\frac{\mathrm{a}}{2}=\frac{2.08\times {10}^{-10}}{2}\mathrm{m}=1.04\times {10}^{-10}\mathrm{m}\\ \approx 1\times {10}^{-10}\mathrm{m}\end{array}$